Complex Numbers And Quadratic Equations question 149
Question: If $ a,b,c $ and $ u,v,w $ are complex numbers representing the vertices of two triangles such that $ c=(1-r)a+rb $ and $ w=(1-r)u+rv $ , where r is a complex number, then the two triangles
Options:
A) Have the same area
B) Are similar
C) Are congruent
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let the complex number $ a,b,c $ and $ u,v,w $ represent the vertices $ A,B,C $ and $ D,E,F $ of the two triangle $ ABC $ and $ DEF $ respectively. Put $ b-a=r_1{e^{i{\theta_1}}} $ $ c-a=r_2{e^{i{\theta_2}}} $ $ v-u={\rho_1}{e^{i{\varphi_1}}},w-u={\rho_2}{e^{i{\varphi_2}}} $ and $ r=\lambda {e^{i\alpha }} $ Substituting these values in the given relations $ c-a=r(b-a) $ and $ w-u=(v-u)r, $ we have $ r_2{e^{i{\theta_2}}}=\lambda {e^{i\alpha }}r_1{e^{i{\theta_1}}}=\lambda r_1{e^{i(\alpha +{\theta_1})}} $ …….(i) and $ {\rho_2}{e^{i{\varphi_2}}}={\rho_1}{e^{i{\varphi_1}}}\lambda {e^{i\alpha }}=(\lambda {\rho_1}){e^{i({\varphi_1}+\alpha )}} $ …….(ii) Equating moduli and arguments of the complex numbers on both sides (i), we get $ r_2=\lambda r_1,{\theta_2}=\alpha +{\theta_1} $ i.e., $ AC=\lambda AB $ and $ \angle CAB={\theta_2}-{\theta_1}=\alpha $ Similarly from (ii), we shall get $ DF=\lambda DE $ and $ \angle FDE={\varphi_2}-{\varphi_1}=\alpha $ Thus we get $ \frac{AC}{DF}=\frac{AB}{DE} $ and $ \angle CAB=\angle FDE $ Hence the triangle $ ABC $ and $ DEF $ are similar.
BETA
